Below is the source code I wrote, and the infixToPostfix function converts the median expression in the main function into a posterior expression, and the evalPostfix function calculates the converted posterior expression. The algorithm that I convert the median expression to the posterior expression is
As I know, when I print out the results of the infixToPostfix I made, I ask because I can't find the answer in my head even if I compile it several times in my head, "Only the numbers that are operands are output and the operators are not output." The result after converting to the posterior equation I thought is 25-34*1-+93/-, and you don't have to worry about the evalPostfix function.
#include<stdio.h>
void infixToPostfix(const char*);
int evalPostfix(const char*);
char stack[50];
char stack_result[50];
int top = -1;
char pop(char* a) {
return a[top--];
}
void push(char* a, char data) {
a[++top] = data;
}
void infixToPostfix(const char* c) {
for (int i = 0; c[i] != NULL; i++) {
if (c[i] == '(') { // left bracket
continue;
}
else if (c[i] == '') { // right parenthesis
char x = pop(stack);
printf("%c", x);
push(stack_result, x);
}
else if (c[i] >= 48 &&c[i] <=57) { // operand
printf("%c", c[i]);
push(stack_result, c[i]);
}
else { // operator
push(stack, c[i]);
}
}
while (top > -1) {
pop(stack);
}
}
int evalPostfix(const char* ch) {
for (int i = 0; ch[i] != NULL; i++) {
if (ch[i] >= 48 &&ch[i] <=57) { // operand
push(stack, ch[i]);
}
else { // operator
int opr2 = pop(stack);
int opr1 = pop(stack);
switch (ch[i]) {
case '+':
push(stack, opr1 + opr2);
break;
case '-':
push(stack, opr1 - opr2);
break;
case '*':
push(stack, opr1 * opr2);
break;
case '/':
push(stack, opr1 / opr2);
break;
}
}
}
return pop(stack);
}
void main() {
int result;
const char* express = "((2-5)+((3*4)-1)-(9/3))";
printf ("median notation: %s", express);
printf ("backward notation: ");
infixToPostfix(express);
result = evalPostfix(express);
printf("\n\nOperation Results => %d", result);
getchar();
}
Below is the output.
Medium notation: (2-5)+(3*4)-1)-(9/3)) Rear notation: 2534 1993
Calculation Results => 0
When I print out the results of the infixToPostfix I made, "Only the numbers that are operands are output, and the operators are not output."
push() and pop() for stack_result stack in the right bracket and operand processing area.
push() and pop() use top variables, so stack_result and stack will share top variables.
As a result, changing one stack also causes problems because the other stack also changes.
Comment on push() and pop() for stack_result in infixToPostfix().
To resolve the problem, you must separate the top variables for each stack.
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